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OpenStudy (anonymous):

Solve for cos2a=(3^1/2)/2

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myininaya (myininaya):

cos(pi/6)=sqrt{3}/2 and cos(11pi/6)=sqrt{3}/2 so cos(2npi+pi/6)=sqrt{3}/2 and cos(2npi+11pi/6)=sqrt{3}/2 for n=0,1,2,... so 2npi+pi/6=2a and 2npi+11pi/6=2a solve both for a

myininaya (myininaya):

1/2(2npi+pi/6)=a 1/2(2npi+11pi/6)=a

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